Answer
a) Section 8
b) Section 5
Work Step by Step
We are given initial angular velocity $\omega_0$, final angular velocity $\omega=0$ and deceleration $\alpha=-0.2rev/s^2$
We will find the wheel's angular displacement, using the following equation: $$\theta=\frac{\omega^2-\omega_0^2}{2\alpha}=\frac{\omega_0^2}{0.4} rev$$
a) For $\omega_0=+1.2rev/s$, we have $\theta=+3.6rev$
This means the wheel will do 3 turns plus $0.6$ turns. Since 1 turn equals $360^o$, 0.6 turns equals $216^o$. Since $\frac{216^o}{30^o}=7.2$, the marker has passed section 7 and will point in section 8.
b) For $\omega_0=+1.47rev/s$, we have $\theta=+5.4rev$
This means the wheel will do 5 turns plus $0.4$ turns. Since 1 turn equals $360^o$, 0.4 turns equals $144^o$. Since $\frac{144^o}{30^o}=4.8$, the marker has passed section 4 and will point in section 5.
